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If ABCD is a square, what are the coordinates of C? [#permalink]

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22 Mar 2014, 14:00

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If ABCD is a square, what are the coordinates of C?

A. \(\sqrt{3} , \sqrt{3}\)

B. \(\sqrt{3}, 1 + \sqrt{3}\)

C. \(2 \sqrt{3} , \sqrt{3}\)

D. \(1 + \sqrt{3} , \sqrt{3}\)

E. \(\sqrt{3} , 2 \sqrt{3}\)

Guys - I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees?

If ABCD is a square, what are the coordinates of C?

A. \(\sqrt{3} , \sqrt{3}\)

B. \(\sqrt{3}, 1 + \sqrt{3}\)

C. \(2 \sqrt{3} , \sqrt{3}\)

D. \(1 + \sqrt{3} , \sqrt{3}\)

E. \(\sqrt{3} , 2 \sqrt{3}\)

Guys - I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees?

That's because the straight line is 180 degrees (30+90+60) and not 150 degrees (30+90+30).
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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28 Jul 2014, 19:26

If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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28 Jul 2014, 19:53

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GeorgeA023 wrote:

If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

Please refer the screenshot

It is a thumb rule; if a right triangle has angles 30 - 60 - 90; sides correspondingly opposite would be \(1 - \sqrt{3} - 2\)

If ABCD is a square, what are the coordinates of C?

A. \(\sqrt{3} , \sqrt{3}\)

B. \(\sqrt{3}, 1 + \sqrt{3}\)

C. \(2 \sqrt{3} , \sqrt{3}\)

D. \(1 + \sqrt{3} , \sqrt{3}\)

E. \(\sqrt{3} , 2 \sqrt{3}\)

There are many other ways of solving this question:

One method is to note that the sides of 30-60-90 triangle are in the ratio 1:\(\sqrt{3}\):2. Since OA has length 1, OD has length \(\sqrt{3}\). Because the square is tilted, x coordinate of C will be more than x coordinate of D. So options (A), (B) and (E) are out. Also, the square is slightly titled so the x coordinate of C will not be twice of x coordinate of D. Hence option (C) is not possible either. Answer must be (D).

We see that coordinates of D are (\(\sqrt{3}\), 0) because of the 30-60-90 triangle. Now imagine that the x axis is turned 90 degrees at D. So the y coordinate of C will be \(\sqrt{3}\). Also since OA is of length 1, when you rotate it by 90 degrees, the x coordinate of C is 1 more than the x coordinate of C i.e. it is \(\sqrt{3} + 1\).
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Is this a special pythagorean triangle like the ones with sides 3,4,5 and 5,12,13?

Pythagorean triplets give you the ratio of sides where all three lengths of sides are integers - the most common one is 3, 4, 5.

This is the ratio of sides in case the right triangles are special triangles:

Triangle with angles 30-60-90 (occur often) - Here ratio of sides is 1: root 3: 2 (Not a pythagorean triplet since all sides are not integers) Isosceles right triangle i.e. angles are 45-45-90 - Here ratio of sides is 1:1:root 2 (Not a pythagorean triplet since all sides are not integers)
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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18 Jul 2015, 03:18

PareshGmat wrote:

GeorgeA023 wrote:

If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

Please refer the screenshot

It is a thumb rule; if a right triangle has angles 30 - 60 - 90; sides correspondingly opposite would be \(1 - \sqrt{3} - 2\)

Can anyone Pls. explain , how did we get 1 for one of the sides of the blue triangle. I solved till that point and could not proceed.

Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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18 Jul 2015, 03:40

robinpallickal wrote:

Can anyone Pls. explain , how did we get 1 for one of the sides of the blue triangle. I solved till that point and could not proceed.

As mentioned above, after you get the angles as 30-60-90 in the blue triangle, you need to remember that the sides for a 30-60-90 triangle are in the ratio \(1:\sqrt{3}:2\) with '1' side opposite to 30 degree angle. Thus you get 1 for that length.

Another way is by trigonometry: in the blue triangle, you know that the hypotenuse =2

Thus in this triangle, cos (60) = base / hypotenuse ---> 1/2 = base / 2 ---> base = 1 (as cos(60) = 0.5)

If ABCD is a square, what are the coordinates of C?

A. \(\sqrt{3} , \sqrt{3}\)

B. \(\sqrt{3}, 1 + \sqrt{3}\)

C. \(2 \sqrt{3} , \sqrt{3}\)

D. \(1 + \sqrt{3} , \sqrt{3}\)

E. \(\sqrt{3} , 2 \sqrt{3}\)

Guys - I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees?

An explanation using property of slopes!

Product of the slopes of Perpendicular lines = -1

\(m_1*m_2 = -1\) Since AO = 1 therefore OD = √3 [30-60-90 Triangle ratio] Let, Slope of Line AD, \(m_1 = -1/√3\) i.e. Slope of Line DC, \(m_2 = √3\) in Triangle i.e. Ratio of Y-Co-ordinate of C to Horizontal Distance of CD on X-Axis = √3/1

i.e. x Co-ordinate of C = OD+1= √3+1

i.e. Co-ordinates of C = ((√3+1),√3)

i.e. Option D
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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22 Nov 2015, 10:36

since we have 30-60-90 triangle, we can deduct that the sides of the small triangle with the base on the x axis is 1-sqrt(3)-2 which makes the length of a side of the square = 2. since all the sides are equal, we can draw a line from point c to the x-axis to form a 30-60-90 triangle. since we have a similar triangle with the first one, we see that the hypotenuse is the same = 2. which makes that the point on y axis is sqrt(3)

now, only points A, C, and D have y coordinate of point c sqrt(3). B and E are thus eliminated. now, from the origin to the point D, we have a distance of sqrt(3). The distance from point D to the line perpendicular to the x-axis which form a 30-60-90 triangle is 1. Thus, the x-coordinate of the point C must be 1+sqrt(3)

Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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11 Apr 2016, 20:11

PareshGmat wrote:

GeorgeA023 wrote:

If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

Please refer the screenshot

It is a thumb rule; if a right triangle has angles 30 - 60 - 90; sides correspondingly opposite would be \(1 - \sqrt{3} - 2\)

For Red Triangle cos 60 =adjancent side/hyp 1/2 =1/h h=2

Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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30 Nov 2017, 03:24

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